Minimum force required to rotate a lamina

[PhysicsBoi1908]

Homework Statement

A uniform right angled triangular lamina is placed on a horizontal floor, which is not frictionless. One of the acute angles of lamina is theta. If F_A and F_B are the minimum forces required to rotate the lamina about stationary vertical axes through A and B respectively, then find the minimum force required to rotate the lamina about a stationary vertical axis passing through C.

Relevant Equations

τ=Iα

When the lamina rotates about A, F_A must act on B (because it is the farthest away) perpendicular to AB (so that all of F_A contributes to rotation).
Same argument is valid for rotation of lamina about B as well.

Having noted that, I tried two approaches:

Approach 1-



If I assume that the lamina has mass m, then maximum static friction becomes μmg. F_C must act on A such that it is perpendicular to AC. Then I just have to equate F_Clcosθ=∫dfr, where l is the length of the hypotenuse.
I can find df by writing dfr=dmαr
where F_Clcosθ=I_Cα, where I_C is the moment of inertia of the lamina about C.

There are a lot of problems with this approach:
I don't know I_C, I couldn't calculate it.
If I evaluated F_C this way, then I would have to repeat the process for F_A and F_B as well, which would make the solution very lengthy.

Approach 2-



While the last approach was, according to me, theoretically correct, I can't assure that for this one.
I argue that the minimum force required to turn A and B (and thus the lamina) about C must be the minimum force required to rotate lamina (and thus C) about A and B.
Then, the torque due to these forces about C must equal F_Clcosθ.
This approach dues give an answer, albeit the wrong one.

I think I know why the answer comes out to be incorrect. If F_{A or B} is enough to cause rotation of A or B about C, then it can effectively rotate the lamina, and thus the answer must either by F_{A or B} or something less than that.
Indeed, the correct answer is less than what I get from the above approach.

At this point, I have no more ideas.
(Note for approach one, I did find some videos online which derive moment of inertia of a triangular lamina, using "area moment" or something like that. But I would prefer to solve this question using high school physics, as the problem book expects me to.

 

[etotheipi]

Here's an idea, it could well be wrong... I think we can assume that any frictional force acts at the centroid. Tentative justification: if we assume that a uniform frictional force per unit mass $\vec{f}$ acts on the lamina (which has a uniform mass density $\mu$), then the net torque of this friction about the centre of mass is$$\iint_S \vec{r} \times \vec{f} \mu dA = \left(\iint_S \mu \vec{r} dA \right) \times \vec{f} = \vec{0} \times \vec{f} = \vec{0}$$due to the definition of the centre of mass. So the friction on the lamina can be replaced equivalently with a single force through the centroid.

So in each case, it seems to me you will have an applied force (say $\vec{P}$) which for the limiting case, as you have noted, must act at the point furthest from the axis under consideration in a direction perpendicular to the the line between them. There would also be a frictional force $\vec{F}_F$ that acts through the centre of mass, of magnitude $\mu mg$. Then there is also the force from the hinge.

I'll check this tomorrow, but does that sort of agree with what you did?

 

[PhysicsBoi1908]

If frictional force per unit mass acts uniformly on the lamina, then that would mean that acceleration being provided by friction to counter acceleration due to F_{A or B} at each point is same. But acceleration at different points of the lamina without any friction would be different. The angular acceleration would be same.

If, say, there is an angular acceleration $\alpha$ about A due to F_A, the friction at some point at distance $r$ away from the point A must provide an acceleration of $\alpha r$.
So think friction can't be uniform.

 

[etotheipi]

If fractional force per unit mass acts uniformly on the lamina, then that would mean that acceleration being provided by friction to counter acceleration due to F_{A or B} at each point is same. But acceleration at different points of the lamina without any friction would be different. The angular acceleration would be same.

As far as I can tell there are, in each trial, three forces on the lamina. The applied force $\vec{P}$, the frictional force $\vec{F}$ (replaced equivalently through the centroid) and the contact force from the hinge $\vec{N}$. With this you have the equation for static equilibrium$$\vec{N} + \vec{F} + \vec{P} = \vec{0}$$You also have $\alpha=0$ about the axis through the hinge, so the torque about that axis is also zero, which you could write as$$P = \mu mg \frac{r_F}{r_P}$$where $r_P$ and $r_F$ are the perpendicular distances from the hinge to the lines of action of those forces respectively. I would try to write this torque equation for each of the three cases.

 

[PhysicsBoi1908]

I actually thought about replacing the frictional force with an equivalent concentrated at a single point earlier, but that would again take approach 1 which I couldn't work out.

 

[etotheipi]

I actually thought about replacing the frictional force with an equivalent concentrated at a single point earlier, but that would again take approach 1 which I couldn't work out.

If we do make the assumption of uniform distribution of friction (which seems fairly reasonable to me anyway...), then for the first case (hinge at A) I would write something like$$F_A = \mu mg \frac{d}{\sqrt{x^2 + y^2}}$$where $x$ and $y$ are the known lengths of the legs of the triangle, and $d$ is the distance between $A$ and the centroid (which I am too lazy to calculate at the moment ). Then I would write similar equations for $B$ and $C$, and use those three equations to eliminate $x$ and $y$ to get $F_C$ in terms of $F_B$ and $F_A$.

But this would all rely on the uniform friction assumption... although it is at least a start

 

[PhysicsBoi1908]

With reference to post #3 and #4:

$$\iint_S \vec{r} \times \vec{f} \mu dA = \iint_S \mu \vec{r} \times \vec{\alpha} \times \vec{r_A} dA = \iint_S \mu \vec{r} \times \vec{\alpha} \times \left( \vec{r_{COM}}+\vec{r} \right) dA =\not = 0$$

$\vec{r}$ is position of element from COM, $\vec{r}_A$ is position of element from hinge A, and $\vec{\alpha}$ is angular acceleration provided by torque

But yes, it is at least a start.

 

[etotheipi]

$$\iint_S \vec{r} \times \vec{f} \mu dA = \iint_S \mu \vec{r} \times \vec{\alpha} \times \left( \vec{r_a}-\vec{r} \right) dA \not = 0$$

If you want to consider an individual mass element in the rigid body, do not neglect internal constraint forces...

 

[PhysicsBoi1908]

I have substituted $\vec{f}$ with $\vec{\alpha}\times\vec{r_A}$ where $\vec{r_A}$ can be written as a sum of $\vec{r}_{COM}+\vec{r}$.
$\vec{\alpha}$ is angular acceleration provided by torque $\vec{l) \times \vec{F_A}$.

I agree with your second statement. I have also corrected post #7.

 

[PhysicsBoi1908]

I am making too many typing mistakes right now. There have been more edits than comments. I guess I will go to sleep.

 

[etotheipi]

I have substituted $\vec{f}$ with $\vec{\alpha}\times\vec{r_A}$ where $\vec{r_A}$ can be written as a sum of r→+rCOM→.
α→ is angular acceleration provided by torque $\vec{l)\ times\vec{F_A}$.

You are considering a mass element on the lamina. The frictional force on this is $\mu dA \vec{f}$. But there will also be internal constraint forces, which we must lump into another term $\vec{F}_c$.

We could look at the forces on this element and write $$\mu dA \vec{f} + \vec{F}_c = \mu dA \vec{a}$$ where $\vec{a} = \alpha \times \vec{r}_A$, but neglecting the internal constraint forces would be a mistake.

But in any case, my derivation was to show that a uniform friction distribution is equivalent to a single force through the centre of mass, and nothing more

 

[PhysicsBoi1908]

Internal constraint forces? I have never heard of this before... Can you please elaborate?
I mean, I have used constraints to form equations, like for when a box falls down a wedge.

 

[etotheipi]

Internal constraint forces? I have never heard of this before... Can you please elaborate?

A rigid body you can think of as a collection of individual mass elements/particles. If you have a rigid box, for instance, and you apply a force to one side, you will notice that all of the particles in the box accelerate in order to satisfy the rigidity condition.

However, the external force only acts on the particles on the one edge of the box where you apply the force. What happens is that every particle in the rigid body exerts forces on other particles in the rigid body that constrain them to all accelerate in a coordinated way.

Have a look here :
https://en.wikipedia.org/wiki/Rigid_body_dynamics#Rigid_system_of_particles

 

[PhysicsBoi1908]

Hmm... the name did give me that vibe. But can we calculate it? Also, will it be uniform?

 

[PhysicsBoi1908]

I think I got it. The non-uniformity is due to the internal constraint forces. But how can we be completely certain that there would be no non-uniformity, as you claim, in $\mu dA \vec{f}$?

 

[etotheipi]

Hmm... the name did give me that vibe. But can we calculate it? Also, will it be uniform?

Some of the time, yes. You can imagine taking a slice through a body, and you can calculate the internal forces between the two resulting parts. You can use this sort of reasoning when dealing with things like beams, for example. You could also look at things like the traction vector and the stress tensor of continuum mechanics. But this is probably not too relevant here.

I think I got it. The non-uniformity is due to the internal constraint forces. But how can we be completely certain that there would be no non-uniformity, as you claim, in $\mu dA \vec{f}$?

It is a good point. I started off by assuming the friction along the lower face of the lamina was uniform for each possible area element, and that is if anything the assumption that could be right or wrong.

If we take it to be true, then it would seem that we can replace friction equivalently with a force through the CM. This proof is identical to the one we'd use to show gravity (or indeed any uniform body force) acts effectively through the CM.

And when we have that, we can consider the equations of motion for the rigid body as a whole. So we wouldn't ever need to worry about the internal forces in the lamina in this setup, which would all cancel pairwise.

 

[PhysicsBoi1908]

I have never seen this before. Thank you for showing this.

What if I consider a radial element in the lamina, then the magnitude of acceleration of each point will be same for mass elements in it.
Since the radial element is a system of mass elements comprising it, I need not consider the internal constraint forces between them.
The only internal constraint forces will be from the radial element before and after the element in question.

Assume that frictional force is uniform over the lamina. Now since each point on the radial element has the same magnitude of acceleration, the internal constraint forces must have uniformly accelerated the points, because the only other force is the frictional force, which as per our assumption is uniform.
But this concludes total uniform acceleration over the lamina, which we know isn't true.
Thus, by contradiction, frictional forces must be non-uniform.
I could very well be wrong though.

edit: it doesn't. I just realized that the internal constraint forces between radial elements can differ as well. It just shows that these forces will be uniform for a radial element. Stupid me.

 

[etotheipi]

I wouldn't worry too much about the internal forces, especially not for a lamina. You just need to keep in the back of your mind that they are there, in case you ever try to think about the forces on a small piece.

Really I have no idea if friction is uniform across the surface. That was just an assumption I made, on which the whole rest of what I said lies. It could well be incorrect! But that was the only way I could see how to solve the problem. Maybe it is OK as a first approximation.

Perhaps someone else will have a better idea

 

[PhysicsBoi1908]

I did calculations assuming friction was concentrated at the COM.
Here are the pics:



The answer is incorrect. This approach also allows to write $F_C$ in terms of $F_A$ and $F_B$ individually (since $\theta$ is actually given), which shouldn't be possible.

 

[etotheipi]

In that case I'm out of ideas also. I guess you could try postulating that friction exerts an unknown maximum torque $\tau_F$ which is the same in all three scenarios and see if that gets the correct answer.

 

[PhysicsBoi1908]

That doesn't work either, I tried it just now. Also, the torque due to friction in case of A would be greater than in the case of B and C. Thank you for your help.

 

[etotheipi]

Also, the torque due to friction in case of A would be greater than in the case of B and C.

But we don't know this unless we know something to do with the spatial distribution of the friction force.

Perhaps someone will come along and have a better idea, but at this point I am afraid that I cannot read the author's mind .

 

[PhysicsBoi1908]

I think the author doesn't want us to think about internal forces at all when we consider an element of the lamina. Then for a radial element, the torque due to friction would remain constant. I'll think about this and post if something comes up.

 

[etotheipi]

I think the author doesn't want us to think about internal forces at all when we consider an element of the lamina.

Well if you consider an element of the lamina, you must include the internal forces otherwise it will definitely be wrong. That is something that is true for certain.

But usually you never consider applying dynamical equations to an individual element of the lamina or rigid body, at least not in cases like this. So I think even mentioning internal forces is going down the wrong path.

I am presuming the author obtained his/her answer by writing some equilibrium equations for the lamina as a whole, having made certain assumptions about the action of friction. But without knowing how we are supposed to treat friction, we are just guessing

 

[haruspex]

if we assume that a uniform frictional force per unit mass $\vec{f}$ acts on the lamina (which has a uniform mass density $\mu$), then the net torque of this friction about the centre of mass is$$\iint_S \vec{r} \times \vec{f} \mu dA = \left(\iint_S \mu \vec{r} dA \right) \times \vec{f} = \vec{0} \times \vec{f} = \vec{0}$$

The magnitude per unit area will be constant but not the direction.
The torque will be $\int |\vec r|\mu\rho g.dA$, somewhat of a messy integral over a triangle.

 

[etotheipi]

The magnitude per unit area will be constant but not the direction.
The torque will be $\int |\vec r|\mu\rho g.dA$, somewhat of a messy integral over a triangle.

Right, so in that case we are saying that the direction of friction is perpendicular to the lever arm? That would make sense. The problem does become a little more difficult

 

[haruspex]

Right, so in that case we are saying that the direction of friction is perpendicular to the lever arm? That would make sense. The problem does become a little more difficult

I did wonder whether there was a way to avoid the integral by considering the relationship between the distances from an element to the three vertices, but that looks messy too.
Found an interesting determinant equation at https://mathoverflow.net/questions/347367/distance-between-point-inside-a-triangle-and-its-vertices, but plugging in that it is a right angled triangle doesn't seem to simplify it.
Also, for the axes A and B, the torque arm of the applied force is the hypotenuse, but for axis C it is the longer of the other two sides.

 

[PhysicsBoi1908]

Force will be applied on A. AC is the larger side, since $\theta$ is acute.

 

[haruspex]

I see a way.
Drop a perpendicular from C to meet AB at D. Consider the torques required to rotate the two smaller triangles about C.

 

[PhysicsBoi1908]

This is very clever. Thank you very much.

I have a doubt though, in post #25 you concluded that the magnitude of friction per unit area will be constant, how did you arrive at that? More generally, how do we conclude that magnitude of friction is constant in any scenario?

 

[haruspex]

how do we conclude that magnitude of friction is constant

The weight per unit area is constant. When the lamina moves, all parts go into kinetic friction, so when it is about to move all parts must be at max static friction.

 

[PhysicsBoi1908]

Ah! Thank you very much.

 

[etotheipi]

I see a way.
Drop a perpendicular from C to meet AB at D. Consider the torques required to rotate the two smaller triangles about C.

It is very clever! If you don't explicitly cut the lamina there will be an internal force between the two parts. The result is that the torque you have to apply at A/B is slightly more than calculated and the torque you have to apply at B/A is slightly less than expected. But these two discrepancies will cancel exactly, and their sum will just be the same as if the lamina were not actually connected, except at the hinge (i.e. no internal force between the two parts).